Solving Hard Problems in Physics: Finding Minimum Force and Acceleration

  • Thread starter Thread starter allblonde
  • Start date Start date
  • Tags Tags
    Hard
AI Thread Summary
The discussion centers on solving physics problems involving forces on an incline and multiple blocks on a frictionless surface. For the crate on a 35-degree incline, the minimum force needed to prevent sliding is calculated to be 32.2 N, but there is confusion about resolving vectors correctly. Participants emphasize the importance of understanding the forces acting on the crate, including gravitational and frictional forces, and how they differ on an incline compared to a flat surface. In the second problem, the acceleration of three blocks under a horizontal force is discussed, with suggestions to treat it as a center of mass problem. Accurate force resolution is crucial for both scenarios to arrive at the correct answers.
allblonde
Messages
7
Reaction score
0
:cry: some other problems if anyone has any ideas

1. the coefficient of static friction between the 3.00 kg crate and the 35 degree incline is .300. What is the magnitude of the minimum force, F, that must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

: so i know the answer is 32.2 N but I am not sure how to get it , do i need to resovle the vectors of gravity(weight)

2.Three blocks are in contact with each other on a frictionless horizantal surface. A horizantal force, F, is applied to m1 (the first box). For this problem, m1=2kg, m2=3kg, m3=4kg, and F= 180N to the right.
a. find the acceleration of the blocks.
b. find the resultant force on each block.
c. find the magnitude of the contact forces.
 
Physics news on Phys.org
i know:
force perpendicular (y) is the force applied*the sin of theta
force parallel (x) is the force applied*the cos of theta
the force due to gravity is the mass*gravity
Fn+Fp-Fg= 0 because the crate is not floating
Fn=-Fp+Fg
Fk=coefficient of friction*Fn
Force perpendicular-Fk=m(a of x)

this is easy to understand and solve when the crate is on a flat horizantal surface, but when i try to plug these things into a problem with an incline, nothing comes out right i don't know if I am resolving wrong or what
 
allblonde said:
i know:
force perpendicular (y) is the force applied*the sin of theta
force parallel (x) is the force applied*the cos of theta
the force due to gravity is the mass*gravity
Fn+Fp-Fg= 0 because the crate is not floating
Fn=-Fp+Fg
Fk=coefficient of friction*Fn
Force perpendicular-Fk=m(a of x)

I think you're forces are incorrect for the first problem. The vertical force is perpendicular to the surface of the incline so there should not be any \theta dependence. I have for my summation of forces
F_y: -F+F_n-mg\cos \theta
and
F_x: -mg\sin \theta +\mu_{s}F_n=0

The second problem is just doing the same thing three times or think of this as a center of mass problem. Good luck!
 
Last edited:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top